Trigo Prob to prove

  • Thread starter ron_jay
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  • #1
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Homework Statement



if: [tex] (1+\sqrt{1+a})\tan{\alpha} = (1+\sqrt{1-a}) [/tex]

Prove that: [tex] \sin{4 \alpha} = a [/tex]

Homework Equations



[tex] \cos{2\alpha} = 1-2\sin^2{\alpha} [/tex]

[tex] \tan{\alpha} = \sqrt{\frac{1-\cos{2\alpha}}{1+\cos{2\alpha}}}[/tex]

The Attempt at a Solution



We probably have to simplify to express [tex] \tan{\alpha} [/tex] in terms of [tex]\sin{4\alpha}[/tex] and hence need to use the equations given above. Trying componendo and dividendo did no good as it got too complicated.

[tex]\sin{\alpha}+\sin{\alpha}\sqrt{1+a} = \cos{\alpha}+\cos{\alpha}\sqrt{1-a}[/tex]

[tex](\sin{\alpha}-\cos{\alpha})^2 = (\cos{\alpha}\sqrt{1-a}-\sin{\alpha}\sqrt{1+a})^2 [/tex]

[tex]1-\sin{2\alpha}=\cos^2{\alpha}(1-a)+\sin^2{\alpha}(1+a)-\sin{2\alpha}\sqrt{1-a^2}[/tex]

Though I could simplify after this, I cannot understand how to get [tex]\sin{4\alpha} [/tex] out of it. What is looking even more difficult is how to remove the [tex] \sqrt{1-a^2} [/tex] term.

I would be grateful if somebody could guide me in solving this.
 

Answers and Replies

  • #2
45
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Wait, is the [tex] \sqrt{1-a^2} [/tex] inside of your last sine term or is it a scalar multiplying your last sine term?
 
  • #3
81
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No, it is a scalar multiplying the last sine term.
 
Last edited:
  • #4
81
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Looks like I got it; beginning from where I left,

[tex]1-\sin{2\alpha} = \cos^2{\alpha-a\cos^2{\alpha}+\sin^2{\alpha}+a\sin^2{\alpha}-(\sin{2\alpha})(\sqrt{1-a^2})[/tex]

-After taking 'a' common...

[tex]Since, \cos^2{\alpha}+\sin^2{\alpha}=1[/tex]

[tex]\Rightarrow\sin{2\alpha}=a\cos{2\alpha}+(\sin{2\alpha})(\sqrt{1-a^2})[/tex]

[tex]\Rightarrow(\sin{2\alpha}-a\cos{2\alpha})^2=(\sqrt{1-a^2}.\sin{2\alpha})^2[/tex]

[tex]\Rightarrow\sin^2{2\alpha}+a^2\cos^2{2\alpha}-a\sin{4\alpha}=\sin^2{2\alpha}-a^2\sin^2{2\alpha}[/tex]

the term [tex]sin^2{2\alpha}[/tex] gets cancelled on both sides and taking 'a' common we get>>

[tex]\Rightarrow \sin{4\alpha}=a[/tex]

VOILA!
 

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